Evaluate Tangent of Difference of Inverse Trigonometric Functions

To evaluate the given expression considering only the principal values of the inverse trigonometric functions:

$\tan \left( \sin^{-1} \left( \frac{3}{5} \right) - 2 \cos^{-1} \left( \frac{2}{\sqrt{5}} \right) \right)$

Let: $\alpha = \sin^{-1} \left( \frac{3}{5} \right) \quad \text{and} \quad \beta = \cos^{-1} \left( \frac{2}{\sqrt{5}} \right)$

Since we are using principal values, both $\alpha$ and $\beta$ lie in the first quadrant, i.e., $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$.

From $\alpha = \sin^{-1} \left( \frac{3}{5} \right)$, we have: $\sin \alpha = \frac{3}{5}$ Using the Pythagorean identity, $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \frac{4}{5}$. Thus: $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$

From $\beta = \cos^{-1} \left( \frac{2}{\sqrt{5}} \right)$, we have: $\cos \beta = \frac{2}{\sqrt{5}}$ Using the Pythagorean identity, $\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \frac{4}{5}} = \frac{1}{\sqrt{5}}$. Thus: $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$

Next, we compute $\tan(2\beta)$ using the double-angle formula for tangent: $\tan(2\beta) = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2 \left( \frac{1}{2} \right)}{1 - \left( \frac{1}{2} \right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

Now, using the angle difference formula for tangent, we evaluate $\tan(\alpha - 2\beta)$: $\tan(\alpha - 2\beta) = \frac{\tan \alpha - \tan(2\beta)}{1 + \tan \alpha \tan(2\beta)}$

Substituting the values of $\tan \alpha = \frac{3}{4}$ and $\tan(2\beta) = \frac{4}{3}$: $\tan(\alpha - 2\beta) = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \left( \frac{3}{4} \right) \left( \frac{4}{3} \right)} = \frac{\frac{9 - 16}{12}}{1 + 1} = \frac{-\frac{7}{12}}{2} = -\frac{7}{24}$

Thus, the value of the expression is $\frac{-7}{24}$, which corresponds to Option (B).